9.4 Properties of Rotations POD: 2 3 5 −7 −1 −2 3 −4 2. 2 3 5 −7 + −1 −2 3 −4 3. 2 3 5 −7 − −1 −2 3 −4

Obj: Rotate figures about a point Rotation – a transformation is which a figure is turned about a fixed point. Center of rotation – the fixed point in which a figure is turned about. Angle of rotation – formed from rays drawn from the center of rotation to a point and its image

If P is the center of rotation then mQPQ’ = the rotation. If Q is the center of rotation, then the image of Q = Q

Graph and draw a rotation 1. Draw a 100 rotation of ABC about P.

Coordinate Rules : When point (a,b) is rotated ccw about the origin… 1. rotation of 90 (a,b) => (-b,a) 2. rotation of 180 (a,b) => (-a, -b) 3. rotation of 270 (a,b) => (b, -a)

Graph using coordinates Graph quadrilateral RSTU with vertices R(1,-1), S(3,-1), T(3,-5) and U(0,-4). Then rotate the quadrilateral 270 about the origin. Graph JKL with vertices J(4,1), K(5,4) and L(7,0). Rotate the triangle 90 about the origin.

Using Matrices Rotation Matrices CCW 90 rotation 180 rotation 0 −1 1 0 −1 0 0 −1 270 rotation 360 rotation 0 1 −1 0 1 0 0 1

Trapezoid EFGH has vertices E(-4, 1), F(-4, 3), G(0, 3), and H(1, 1) Trapezoid EFGH has vertices E(-4, 1), F(-4, 3), G(0, 3), and H(1, 1). Find the image matrix for a 180° rotation of EFGH about the origin. Graph EFGH and its image.

Theorem… Theorem 9.3 Rotation Theorem – A rotation is an isometry

Cases of Theorem 9.3 Page 593 Need to prove the rotation preserves the length of a segment. Consider a segment QR rotated about point P to produce segment Q’R’. Case 1 – R, Q and P are noncollinear. Case 2 – R, Q, and P are collinear. Case 3 – P and R are the same point

Prove it… Find the value of y in the rotation of the quadrilateral.

Find the value of y in the rotation of the quadrilateral. By Theorem 9.3, the rotation is an isometry, so corresponding side lengths are equal. Then 3x = 9, so x = 3. Now set up an equation to solve for y. 4y = 6x-2 4y = 6(3) -2 y = 4